Q 11 :

Energy released when two deuterons (H21) fuse to form a helium nucleus (He42) is:             

(Given: Binding energy per nucleon of H21=1.1MeV and binding energy per nucleon of He42=7.0MeV)                [2025]

  • 8.1 MeV

     

  • 5.9 MeV

     

  • 23.6 MeV

     

  • 26.8 MeV

     

(3)

H2+H21He421

Q=BEProduct-BEReactant

Q=(4×7-4×1.1)MeV=23.6 MeV



Q 12 :

The mass of proton, neutron and helium nucleus are respectively 1.0073 u, 1.0087 u and 4.0015 u. The binding energy of helium nucleus is          [2023]

  • 14.2 MeV

     

  • 28.4 MeV

     

  • 56.8 MeV

     

  • 7.1 MeV

     

(2)

B.E. of Helium=(2mp+2mN-mHe)c2=28.4 MeV



Q 13 :

A92238B90234+D24+Q

In the given nuclear reaction, the approximate amount of energy released will be:              [2023]

[Given, mass of A92238=238.05079×931.5 MeV/c2

mass of B90234=234.04363×931.5 MeV/c2

mass of D24=4.00260×931.5 MeV/c2]

  • 3.82 MeV

     

  • 5.9 MeV

     

  • 2.12 MeV

     

  • 4.25 MeV

     

(4)

Q=(mA-mB-mD)×931.5 MeV

     =(238.05079-234.04363-4.00260)×931.5

 4.25 MeV
  



Q 14 :

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R                   [2023]

Assertion A: The binding energy per nucleon is practically independent of the atomic number for nuclei of mass number in the range 30 to 170.

Reason R: Nuclear force is short ranged.

In the light of the above statements, choose the correct answer from the options given below:

  • Both A and R are true but R is NOT the correct explanation of A

     

  • A is true but R is false

     

  • A is false but R is true

     

  • Both A and R are true and R is the correct explanation of A

     

(4)

Binding energy per nucleon is almost same for nuclei of mass number ranging 30 to 170.



Q 15 :

Nucleus A having Z = 17 and equal number of protons and neutrons has 1.2 MeV binding energy per nucleon. Another nucleus B of Z = 12 has total 26 nucleons and 1.8 MeV binding energy per nucleon. The difference of binding energy of B and A will be ______ MeV.                 [2023]



(6)

For A mass number =34

Total binding energy=1.2×34=40.8 MeV

For B mass number =26

Total binding energy=1.8×26 MeV=46.8 MeV

Difference of BE=6 MeV



Q 16 :

If the binding energy of ground state electron in a hydrogen atom is 13.6 eV, then, the energy required to remove the electron from the second excited state of Li2+ will be: x×10-1 eV. The value of x is _______ .              [2023]



(136)

EH=13.6

ELi2=13.6Z2n2=13.6×99=13.6 eV=136×10-1 eV



Q 17 :

A nucleus with mass number 242 and binding energy per nucleon as 7.6 MeV breaks into two fragment each with mass number 121. If each fragment nucleus has binding energy per nucleon as 8.1 MeV, the total gain in binding energy is ________ MeV.          [2023]



(121)

Initial binding energy=242×7.6 MeV

Final binding energy=121×8.1 MeV+121×8.1 MeV=242×8.1 MeV

Total gain in binding energy=242(8.1-7.6)=121 MeV



Q 18 :

A common example of alpha decay is

            U92238Th90234+He42+Q

Given:  U92238=238.05060 u

            Th90234=234.04360 u

            He24=4.00260 u  and 1u=931.5 MeVe2

The energy released (Q) during the alpha decay of U92238 is ________ MeV.                      [2023]



(4)

Energy released=(Δm)amu×931.5 MeV

                                =(mu-mTh-mHe)amu×931.5 MeV

                                =0.0044×931.5 MeV=4.0986 MeV



Q 19 :

Given below are two statements:

Statement I: For all elements, greater the mass of the nucleus, greater is the binding energy per nucleon.

Statement II: For all elements, nuclei with less binding energy per nucleon transform to nuclei with greater binding energy per nucleon.

In the light of the above statements, choose the correct answer from the options given below:                       [2026]

  • Statement I is true but Statement II is false

     

  • Both Statement I and Statement II are false

     

  • Statement I is false but Statement II is true

     

  • Both Statement I and Statement II are true

     

(3)

Theoretical



Q 20 :

The minimum frequency of photon required to break a particle of mass 15.348 amu into 4 α particles is _______ kHz.

[Mass of He nucleus = 4.002 amu, 1 amu = 1.66×10-27 kg, h=6.6×10-34 J.s and c=3×108 m/s]             [2026]

  • 14.94×1019

     

  • 9×1019

     

  • 9×1020

     

  • 14.94×1020

     

(1)

hν=(4×4.002-15.348)×1.66×10-27×(3×108)2

ν=14.94×1019 kHz